Let $R$ be the ring $\mathbb{Z}[x,y,z]/(x(y^2-4z)-y^2+3z)$. I am trying to prove that the completion of this ring at the ideal $(x)$ is given by $\mathbb{Z}[[x]][y,z]/(x(y^2-4z)-y^2+3z)$. I am not really sure how one would go about do this or if this is even true. Any suggestions are appreciated. Thanks in advance.
From the inverse limit definition of the completion, we can get a map from $\mathbb{Z}[[x]][y,z]/(x(y^2-4z)-y^2+3z)$ to the completion. But essentially I am not sure how to show this if this is an isomorphism.
Let $I$ be an ideal of $\mathbb{Z}[x,y,z]$ and let $\wedge_{x}$ denote the $x$-adic completion. We have \begin{align} &(\mathbb{Z}[x,y,z]/I)^{\wedge_{x}} \stackrel{1}{\simeq} (\mathbb{Z}[x,y,z])^{\wedge_{x}} \otimes_{\mathbb{Z}[x,y,z]} \mathbb{Z}[x,y,z]/I \simeq \mathbb{Z}[y,z][[x]] \otimes_{\mathbb{Z}[x,y,z]} \mathbb{Z}[x,y,z]/I \simeq \mathbb{Z}[y,z][[x]]/I \mathbb{Z}[y,z][[x]] \end{align} where the isomorphism 1 is by part (3) of this. When mixing polynomials with power series, the order matters -- note that the natural map $\mathbb{Z}[[x]][y,z] \to \mathbb{Z}[y,z][[x]]$ is (injective but) not surjective. Note also that the leading coefficient of the polynomial (with respect to $y$) is $x-1$ which is a unit in $\mathbb{Z}[[x]]$ so the desired completion is a finite free $\mathbb{Z}[z][[x]]$-module (of rank $2$).